aleph null = a circular one-lane running track (an endless loop, traversed by discrete steps)
aleph one (assuming continuum hypothesis holds) = a circular running track with a traversable width (an inner to outer lane continuum exists for each step on the above track)
aleph two (assuming continuum hypothesis holds) = all possible combinations of footprints on the above track. (the power set of aleph one)
I don’t get it.
IANAMathematician, and your metaphor sounds kosher to me. But I remember a different metaphor from an essay by Isaac Asimov that I thought was much simpler to understand.
These sets all have the same number of points.
It would take a properly certified rabbi to make it kosher, I believe.
Yes. A one-dimensional line (or loop) is a continuum, whose cardinality is higher than aleph-null. Increasing the dimension does not increase the cardinality.
The same – cubed!
Infinity = Infinity^3
Divide by infinity
1 = 1^3
Works for me!
Short answer: No.
All of the original examples are “countable infinites”, where you could define a “one-to-one and onto” relationship between them and the integers.
The only other infinity I know is the continuous infinity, like the number of points on a line (or line segment, or plane, or n-dimensional space.) If there’s a third, I’d like to learn more about it.
Last I talked to someone who really knew this stuff was a long time ago, but at that time they could define a 1-1/onto correspondence between a line and all points on a plane but one. Since infinity+1 = infinity, they didn’t worry about that one. Of course, if you can map a line to a plane, you can map to any number of dimensions.
My favorite little conundrum about infinity is this one.
An hour before midnight, you put 10 pieces of paper into a hat, and take out one. Repeat this process every time it’s half the time closer to midnight. At midnight, how many pieces of paper are in the hat?
Do the same thing, but this time the pieces of paper are numbered. First you put in 1 to 10, and take out 1. Next time you put in 11 to 20 and take out 2, etc. At midnight, how many pieces of paper are in the hat?
I’m not sure this is true, since, like ultrafilter, I’m not sure I understand the OP’s example.
The set of all sets of points on a line.
For any set, its power set (the set of all subsets of that set) has a higher cardinality than that set itself.
By the way, here’s a good basic explanation of this stuff (different “infinities”).
I assume you’re being facetious, but in case a casual reader gets confused, that’s of course not how it works: x[sup]3[/sup] divided by x is x[sup]2[/sup], not 1[sup]3[/sup].
The powerset (i.e. the set of all subsets) of any set always has a strictly larger cardinality than the original set, so you can just go on constructing ever greater infinities…
And it’s easy to construct a correspondence between the points on a line and the points on a plane, using any plane-filling curve (which has been known since 1890 – I’m guessing your discussion wasn’t that long ago…).
Why make it circular? But fine. If we’re only concerned about the spots on which you might step, and the loop’s length is an irrational multiple of your step size (so you don’t just end up in a finite loop), then, yes, there are aleph_null many spots you might step on.
If your goal is to have a continuum, why not just have the inner to outer continuum, and forget the running track? I mean, this analogy doesn’t add anything to just saying “Hey, suppose we had a continuum.” but complication.
But, yes, if we only look at the spots you might step on, which are again in terms of a discrete step-size forward as above, but where you now have a full continuum of lanes available to you, then the available stepping spots have size aleph_null * continuum = continuum (which is also aleph_one, given the continuum hypothesis).
Sure. This would be the powerset operation, and given the generalized (not just the ordinary) continuum hypothesis, the powerset of aleph_k is indeed aleph_(k + 1). [But why understand the jump from aleph_one to aleph_two so differently than the jump from aleph_zero to aleph_one?]
Also, why make your intuitions so dependent on the generalized continuum hypothesis? Just say beth_one, beth_two, etc., instead of aleph_on, aleph_two, etc., and now GCH doesn’t matter. Though, if your concern is actually to understand the aleph series (where each successive item is the smallest cardinality after the last) rather than the beth series (where each successive item is the power-cardinality of the last one), you’re still left to find a way to do so.
We’re gonna need a bigger hat.
This is a good way to understand, for example, transfinite ordinals [where we have distinct omega, omega^2, and omega^3, in precisely this manner], as well as many other accounts of differing infinities. But it’s not a great way to understand infinite cardinals (which seems to be what the OP’s after), because typically, x, x^2, and x^3 are all equal for an infinite cardinal x.
[For example, if we imagine your lines as broken up into natural number coordinates, then you are creating 1-, 2-, and 3-dimensional grids whose points are specified by a single natural number, an ordered pair of natural numbers, and an ordered triple of natural numbers, respectively. But these all have the same cardinality; you can put pairs of natural numbers in a countably infinite sequence like so: (0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), (3, 0), (2, 1), … . This shows N * N has the same cardinality as N. Then multiplying both by N, we find that N * N * N also has the same cardinality as N * N.
If your lines were thought of as having points indexed by a full continuum, rather than just N, it’s not quite as obvious that they all have the same cardinality, and indeed, but they do, classically. Take the continuum to be indexed by real numbers R, show that R has the same cardinality as 2^N (by showing |R| <= |2^N| and |2^N| <= |R| and then applying the classical Cantor-Schroeder-Bernstein argument), and then observe that 2^N * 2^N = 2^(N + N) = 2^N (using the fact that N + N has the same cardinality as N, as shown by, for example, splitting the naturals into evens and odds, each with the same cardinality as the full naturals).
This argument can break down in non-classical formulations of mathematics (e.g., in the context of intuitionistic logic) where R is not identified with 2^N. Indeed, there’s no continuous correspondence between R and R^2, so any mathematical framework in which all functions of the relevant sort being continuous will consider R and R^2 different sorts of infinities. (Which is a perfectly ok and very ordinary thing to do! It’s just a different thing to look at than other things you might do instead…)].
It may be that Asimov’s essay was on transfinite ordinals; it’s been a very long time since I read the collections.
Plane-filling curves won’t work; the map they induce between a line and a plane is never one-to-one. It’s not possible to have a continuous correspondence between a line and a plane (or any such dimension-changing thing; this is, in some sense, what dimensions are all about).
Rather, to put the points on a line and the points in a plane in correspondence, what one does is first show that the points on a line are in correspondence with 2^N (we can inject R into 2^N by sending a real to the set of rationals less than it, and we can inject 2^N into R by sending a {0, 1}-valued sequence to the value in [0, 1] whose decimal expansion it describes. Thus, R <= 2^N, and 2^N <= R, and so we can classically conclude by the Cantor-Schroeder-Bernstein theorem that R = 2^N).
Then one proves the correspondence for 2^N, where it is easy: 2^N * 2^N = 2^(N + N) = 2^N. [N + N = N by splitting the naturals into evens and odds].
The exact correspondence this produces at the end is rather messy. The basic idea is along the lines of “Take two points, write out their decimal expansions, and intersperse them, to get a single point”, but it ends up not quite exactly that, because one has to get around all the bother about a point potentially having a non-unique decimal expansion (0.9999… = 1.0000…).
Missing words re-instated in bold
Assuming each piece of paper is either placed in the hat to remain there forever or placed in the hat to be at some point removed forever, with the only placings and removings being the ones you’ve noted, then:
It depends on which pieces of paper you remove at each step. Let the f(n)th piece of paper placed in be the nth piece of paper removed; the only constraint is that f(n) is an injective function upper-bounded by 10n. The complement of the range of f(n) can have cardinality anywhere between zero and countably infinite.
No pieces of paper remain
(Of course, we could also ask things like: You have a single piece of paper which you first place into the hat at 11:00, then remove at 11:30, then place in again at 11:45, then remove again at 11:52:30, then place in again… . At midnight, is it in the hat or not?
To which the response, as always, is: Who’s to say? You’re the scenario-specifier. You tell me the rule for how the state of the world at midnight is determined by the state of the world at all times prior to midnight. This can’t be answered except relative to such a rule.)
That makes no sense since none of those sets were well-ordered. In fact, only the first was linearly ordered. The OP makes no sense to me. The first infinite cardinal, aleph-0 is the set of integers (or set of rationals, or…) while the next, aleph-1, may be (this hypothesis is called the continuum hypothesis) the continuum, that is the set of points on the line, also the set of subsets of integers and many other manifestations. A higher cardinal (this might be aleph-2, the next version of the continuum hypothesis) could be thought of as the set of all functions from the line to itself. But the set of continuous functions has the cardinal of the continuum. Past that you can iterate the set of subsets construction as often as you like–through all ordinals. Elephants all the way. When you run out of new ways, you get the first inaccessible cardinal. Then the second, then…maybe a measurable cardinal.
Is it even established that aleph-1 is well-defined? Couldn’t one instead have an infinite hierarchy of infinite cardinals in between aleph-0 and the continuum, such that for any cardinal larger than aleph-0, there’s another cardinal smaller than it but still larger than aleph-0?